# If $y = {x^x}$, prove that $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$ .

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Hint: In this question first we take log both side and simply differentiate expression $y = {x^x}$ with respect to $x$ and find the value of $\dfrac{{dy}}{{dx}}$ & $\dfrac{{{d^2}y}}{{d{x^2}}}$ after that value of $\dfrac{{dy}}{{dx}}$ &$\dfrac{{{d^2}y}}{{d{x^2}}}$ and put left hand side of differential expression $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$.

Let $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$ ……. (1)

Consider $y = {x^x}$

Take log both side

$\log y = \log {x^x}$

Apply log property $\left( {\log {a^b} = b\log a} \right)$

$\log y = x\log x$

Differentiate with respect to $x$

$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x$

$\dfrac{{dy}}{{dx}} = y\left( {1 + \log x} \right)$ ……. (2)

Differentiate equation (2) with respect to $x$

$\dfrac{{{d^2}y}}{{d{x^2}}} = y\left( {0 + \dfrac{1}{x}} \right) + \left( {1 + \log x} \right)\dfrac{{dy}}{{dx}}$

Put value of $\dfrac{{dy}}{{dx}}$ in above expression

$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{y}{x} + y{\left( {1 + \log x} \right)^2}$…….(3)

Value of $\dfrac{{dy}}{{dx}}$ and $\dfrac{{{d^2}y}}{{d{x^2}}}$ from equation 2 and 3 put into equation 1

$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - \dfrac{1}{y}{\left( {y\left( {1 + \log x} \right)} \right)^2} - \dfrac{y}{x} = 0$

$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - y{\left( {1 + \log x} \right)^2} - \dfrac{y}{x} = 0$

Here we can see that after solving left hand side will be zero

So L.H.S=R.H.S

Hence proved

Note: In this question we use log property $\log {a^b} = b\log a$ and also we use some basic differentiation formula like

$

\dfrac{d}{{dx}}\log x = \dfrac{1}{x} \\

\\

$

$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ here $u$ and $v$ are function of real variable $x$ and this formula also known as product rule for derivatives.

$\dfrac{d}{{dx}}x = 1$

Let $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{1}{y}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - \dfrac{y}{x} = 0$ ……. (1)

Consider $y = {x^x}$

Take log both side

$\log y = \log {x^x}$

Apply log property $\left( {\log {a^b} = b\log a} \right)$

$\log y = x\log x$

Differentiate with respect to $x$

$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x$

$\dfrac{{dy}}{{dx}} = y\left( {1 + \log x} \right)$ ……. (2)

Differentiate equation (2) with respect to $x$

$\dfrac{{{d^2}y}}{{d{x^2}}} = y\left( {0 + \dfrac{1}{x}} \right) + \left( {1 + \log x} \right)\dfrac{{dy}}{{dx}}$

Put value of $\dfrac{{dy}}{{dx}}$ in above expression

$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{y}{x} + y{\left( {1 + \log x} \right)^2}$…….(3)

Value of $\dfrac{{dy}}{{dx}}$ and $\dfrac{{{d^2}y}}{{d{x^2}}}$ from equation 2 and 3 put into equation 1

$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - \dfrac{1}{y}{\left( {y\left( {1 + \log x} \right)} \right)^2} - \dfrac{y}{x} = 0$

$\dfrac{y}{x} + y{\left( {1 + \log x} \right)^2} - y{\left( {1 + \log x} \right)^2} - \dfrac{y}{x} = 0$

Here we can see that after solving left hand side will be zero

So L.H.S=R.H.S

Hence proved

Note: In this question we use log property $\log {a^b} = b\log a$ and also we use some basic differentiation formula like

$

\dfrac{d}{{dx}}\log x = \dfrac{1}{x} \\

\\

$

$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ here $u$ and $v$ are function of real variable $x$ and this formula also known as product rule for derivatives.

$\dfrac{d}{{dx}}x = 1$

Last updated date: 19th Sep 2023

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